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irga5000 [103]
3 years ago
14

3. The cell below is currently in Metaphase I. Draw the four resulting cells after the original cellgoes through Meiosis. Write

down the genotype above each of the new cells. (ignore
crossing over).
Biology
1 answer:
snow_tiger [21]3 years ago
3 0

<em>You will find the cell in metaphase and the draw of the gametes in the attached files</em>

Answer:

Gametes: RY, RX, rY, rX

Explanation:

During metaphase I, homologous pairs migrate to the equatorial plane, where they randomly aline with their kinetochores facing opposite poles. The random arrangement of tetrads is different in every cell going through the meiosis process. There is no equal alinement between two cells. When tetrads aline in the equatorial plane, there is no predetermined order for each of the homologous chromosomes of each tetrad to face one of the poles and then migrate to it while separating. Any chromosome of the homologous pair might face any of the poles and then migrate to it. Each of the chromosomes has two possibilities for orientation at the plane.

During metaphase II, when the new haploid cells are formed, the number of variations in each cell is also different and depends on the chromosomes that form that cell. This random order in the equatorial plane is what introduces variation into the gametes. It is almost impossible that two gametes resulting from meiosis will get the same genetic charge.

Diploid cell)  RrXY

Gametes)   RY, rY, RX, rX

The alleles of each of the genes randomly migrate to different poles and combine with the allele of the other gene that migrated to the same pole.

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The count in a bacteria culture was 100 after 20 minutes and 2000 after 40 minutes. Assuming the count grows exponentially, What
Afina-wow [57]

Answer:

Explanation:

Let X be the number of bacteria at time t, and Xo be the initial number of bacteria

Bacteria grows exponentially, the exponential growth model is thus:

X = Xo e^{kt}

[<em>e</em> is the exponential sign]

k is the growth constant = growth rate

At t = 20 minutes, X = 100

At t = 40 minutes, X = 2000

Substituting that into the formula

(i) ...    100 = Xo e^{k20}

(ii) ...    2000 = Xoe^{k40}

<em>Divide (ii) by (i)</em>

\frac{2000}{100} = \frac{Xoe^{40k} } {Xoe^{20k}}

20 = e^{40k-20k} = e^{20k}

<em>[Xo cancels Xo. Division of values raised to a power is done by subtracting their powers]</em>

<em>Take the natural log of both sides</em>

㏑20 = 20k

<em>[taking the ㏑ cancels the exponential]</em>

k = \frac{ln20}{20}

<em>We can now substitute k to solve one of the equations</em>

<em>substituting k in (ii)</em>: 2000 = Xo e\frac{ln 20}{20}.^{40}

2000 = Xo e^{0.15*40}

2000 = Xo e^{6}

<em>Making Xo the subject of the formula</em>

Xo = \frac{2000}{e^{6} }

Xo is approximately 5 cells.

THE DOUBLING TIME

The doubling time is the time it takes for the population to double, so 5 cells become 10 cells

Since Xo = 5

Given that X = Xo e^{kt}

When X=10,

10 = 5 e^{0.15t}

<em>solving for t:</em>

e^{0.15t} = \frac{10}{5}

Take ln of both sides

0.15t = ln 2

t =\frac{ln2}{0.15}

t =4.62 minutes

POPULATION AFTER 65 MINUTES

X = 5 e^{0.15*65}

X = 85770

WHEN WILL THE POPULATION REACH 12000

12000 = 5 e^{0.15t}

e^{0.15t} =240

take ln of both sides

0.15t = ln 2400

t = 51.89 minutes (approximately 52 minutes)

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